What are all the possible rational zeros for #f(x)=4x^4+19x^2-63# and how do you find all zeros?

1 Answer
Aug 26, 2016

Answer:

#x=+-3/2# or #x=+-sqrt(7)i#

Explanation:

#f(x) = 4x^4+19x^2-63#

By the rational roots theorem, any rational zeros of #f(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-63# and #q# a divisor of the coefficient #4# of the leading term.

That means that the only possible rational zeros are:

#+-1/4, +-1/2, +-3/4, +-1, +-3/2, +-7/4, +-9/4, +-3, +-7/2, +-9/2, +-21/4, +-7, +-9, +-21/2, +-63/4, +-21, +-63/2, +-63#

We could try each of these in turn, but there are easier ways to find the zeros of #f(x)#.

Note that #f(x)# is quadratic in #x^2#, so we can use the quadratic formula to find:

#x^2 = (-19+-sqrt(19^2-4(4)(-63)))/(2*4)#

#=(-19+-sqrt(361+1008))/8#

#=(-19+-sqrt(1369))/8#

#=(-19+-37)/8#

i.e. #x^2 = 9/4# or #x^2 = -7#

Since the result is rational, we could have found this using an AC method instead, but at least the quadratic formula gives it to us directly.

Hence #x=+-3/2# or #x=+-sqrt(7)i#